What is 10.5: Vector problems?

Vector models turn geometrical displacements or forces into component equations; equilibrium means that the vector sum of all forces is zero.

How many marks are 10.5 questions worth?

10.5 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 10.5 appeared in?

We have not yet indexed a verified past-paper appearance for 10.5 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

10.5

Use vectors to solve problems in pure mathematics and in context (including forces).

Draft — not yet indexed

Vector problems

Worked answers and methods for 10.5 on Edexcel A-level Maths 9MA0.

Explanation

Vector models turn geometrical displacements or forces into component equations; equilibrium means that the vector sum of all forces is zero.
Choose and state positive coordinate directions, resolve every vector consistently, then equate components or use position-vector relationships.
In a parallelogram with adjacent position vectors $\mathbf{a}$ and $\mathbf{b}$ , the opposite vertex has position vector $\mathbf{a}+\mathbf{b}$ and both diagonals share midpoint $\frac12(\mathbf{a}+\mathbf{b})$ .
A common error is to balance force magnitudes without balancing directions; equal numerical magnitudes do not guarantee equilibrium unless the vector sum is zero.

For the parallelogram with position vectors

\mathbf 0,\mathbf a,\mathbf b,\mathbf a+\mathbf b

, both diagonals share midpoint

(\mathbf a+\mathbf b)/2

Worked example

The points $O,A,B,C$ have position vectors $\mathbf{0},\mathbf{a},\mathbf{b},\mathbf{a}+\mathbf{b}$ respectively. Use vectors to prove that the diagonals $OC$ and $AB$ bisect each other.

1.The midpoint of $OC$ has position vector $\frac12[\mathbf{0}+(\mathbf{a}+\mathbf{b})]=\frac12(\mathbf{a}+\mathbf{b})$ .
2.The midpoint of $AB$ has position vector $\frac12(\mathbf{a}+\mathbf{b})$ as well.
3.Since the two diagonals have the same midpoint, each bisects the other.

Answer: Both diagonals have midpoint position vector $\frac12(\mathbf{a}+\mathbf{b})$ , so they bisect each other.

Common mistakes

Don't assume collinearity from proportional-looking coordinates without finding one common scalar multiplier.
Don't use a diagram alone as proof and never shows that the two candidate midpoints have the same position vector.

Exam tip

In a vector proof, calculate both relevant position vectors and use their equality to justify the geometric conclusion.

Worked practice

Tier 1 · Easy

Two forces acting on a particle are

\mathbf{F}_1=(4,-1)

N and

\mathbf{F}_2=(-2,5)

N. Find the resultant force and its magnitude.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Resultant $(2,4)$ N; magnitude $2\sqrt5$ N	3
Notes
Add the force components: $\mathbf{R}=\mathbf{F}_1+\mathbf{F}_2=(4-2,-1+5)=(2,4)$ N. Its magnitude is $\|\mathbf{R}\|=\sqrt{2^2+4^2}=\sqrt{20}=2\sqrt5$ N.

(3 marks)

Tier 2 · Standard

Two forces acting on a particle are

\mathbf{F}_1=(4,-1)\,\text{N}

and

\mathbf{F}_2=(-2,5)\,\text{N}

. Find the third force required for equilibrium and its magnitude.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$\mathbf{F}_3=(-2,-4)\,\text{N}$ $\|\mathbf{F}_3\|=2\sqrt5\,\text{N}$	3
Notes
For equilibrium the vector sum is zero. The first two forces sum to $(4,-1)+(-2,5)=(2,4)$ , so the third force must be its negative, $(-2,-4)\,\text{N}$ . Its magnitude is $\sqrt{(-2)^2+(-4)^2}=\sqrt{20}=2\sqrt5\,\text{N}$ .

(3 marks)

Tier 3 · Hard

Three vectors sum to the zero vector. Two of them are

\mathbf{a}=(3,4,-2)

and

\mathbf{b}=(-5,1,6)

. Find the third vector

\mathbf{c}

, its exact magnitude and a unit vector in its direction.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$\mathbf{c}=(2,-5,-4)$ $\|\mathbf{c}\|=3\sqrt5$ and a unit direction vector is $\frac{1}{3\sqrt5}(2,-5,-4)$ .	5
Notes
The condition $\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}$ gives $\mathbf{c}=-(\mathbf{a}+\mathbf{b})$ . The first two vectors sum to $(-2,5,4)$ , so $\mathbf{c}=(2,-5,-4)$ . Its magnitude is $\sqrt{2^2+(-5)^2+(-4)^2}=\sqrt{45}=3\sqrt5$ . Dividing the vector by this magnitude gives the unit vector $\frac{1}{3\sqrt5}(2,-5,-4)$ .

(5 marks)

Tier 1 · Easy

Three forces acting on a particle are

(5,-2)

(-1,6)

N and

(-4,-4)

N. Determine whether the particle is in equilibrium, showing the component calculation.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	The particle is in equilibrium because the resultant force is $(0,0)$ N.	2
Notes
The resultant is $(5,-2)+(-1,6)+(-4,-4)=(5-1-4,-2+6-4)=(0,0)$ N. A zero resultant is the condition for equilibrium.

(2 marks)

Tier 2 · Standard

Forces

(7,-4)

N and

(-1,10)

N act on a particle. A third force has the form

\lambda(-1,-1)

N. Find

\lambda

so that the particle is in equilibrium, and find the exact magnitude of the third force.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$\lambda=6$ and the third force has magnitude $6\sqrt2$ N	4
Notes
The first two forces have resultant $(6,6)$ N. Equilibrium requires $(6,6)+\lambda(-1,-1)=(0,0)$ , so $\lambda=6$ and the third force is $(-6,-6)$ N. Its magnitude is $\sqrt{(-6)^2+(-6)^2}=6\sqrt2$ N.

(4 marks)

Tier 3 · Hard

A particle is held in equilibrium by two tensions and its weight. The tensions are

P(\tfrac35,\tfrac45)

N and

Q(-\tfrac5{13},\tfrac{12}{13})

N, where

P>0

and

Q>0

, and the weight is

(0,-56)

N. Find

P

and

Q

, showing both component equations and checking the equilibrium.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$P=25$ N and $Q=39$ N	6
Notes
Horizontal equilibrium gives $3P/5-5Q/13=0$ , so $39P=25Q$ . Vertical equilibrium gives $4P/5+12Q/13-56=0$ . From the first equation $Q=39P/25$ ; substitution into the second gives $4P/5+36P/25=56$ , so $56P/25=56$ and $P=25$ . Hence $Q=39$ . Checking, the two tensions are $(15,20)$ N and $(-15,36)$ N; adding the weight $(0,-56)$ N gives $(0,0)$ N.

(6 marks)

Tier 2 · Standard

A survey vehicle travels

12

km east and

5

km north, followed by

4

km west and

2

km south. Find the vector for the direct return journey, taking east and north as the positive component directions. Find the return distance and the bearing of the return journey, giving the distance to

3

significant figures and the bearing to the nearest degree.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	The direct return vector is $(-8,-3)$ km, the distance is $8.54$ km and the bearing is $249^\circ$ .	5
Notes
The outward displacement is $(12,5)+(-4,-2)=(8,3)$ km, so the direct return vector is $(-8,-3)$ km. Its magnitude is $\sqrt{8^2+3^2}=\sqrt{73}=8.544\ldots$ km, giving $8.54$ km to $3$ significant figures. Measured clockwise from north, the return bearing is $180^\circ+\tan^{-1}(8/3)=249.443\ldots^\circ$ , which is $249^\circ$ to the nearest degree.

(5 marks)

Tier 3 · Hard

A boat moves relative to the water at

12

km h

^{-1}

on a bearing of

030^\circ

. A current of

4

km h

^{-1}

acts due east. Find the boat's ground-velocity vector in east-north component order. Hence find its ground speed, its bearing to the nearest

0.1^\circ

, and the time taken to travel

25

km in the direction of this ground velocity. Give the speed and time to

3

significant figures, using unrounded values in your working.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	The ground-velocity vector is $(10,6\sqrt3)$ km h $^{-1}$ . The ground speed is $14.4$ km h $^{-1}$ , the bearing is $043.9^\circ$ , and the time is $1.73$ h.	6
Notes
The boat's velocity relative to the water has east-north components $(12\sin30^\circ,12\cos30^\circ)=(6,6\sqrt3)$ . Adding the current $(4,0)$ gives ground velocity $(10,6\sqrt3)$ km h $^{-1}$ . Its magnitude is $\sqrt{100+108}=4\sqrt{13}=14.422\ldots$ km h $^{-1}$ . The bearing is $\tan^{-1}[10/(6\sqrt3)]=43.897\ldots^\circ$ , written $043.9^\circ$ . Using the exact speed, the travel time is $25/(4\sqrt{13})=1.7334\ldots$ h, so the required values are $14.4$ km h $^{-1}$ and $1.73$ h to $3$ significant figures.

(6 marks)

Tier 3 · Hard

Two survey drones move in a horizontal coordinate plane, with distances in kilometres and time

t

in hours. Drone

A

starts at

(0,0)

with velocity

(2,1)

, while drone

B

starts at

(10,-2)

with velocity

(-1,3)

. Find the displacement of

B

from

A

at time

t

. Hence find the first time at which the drones are

5

km apart, giving the time to

3

significant figures.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	The displacement of $B$ from $A$ is $(10-3t,-2+2t)$ . The two possible times are $t=\dfrac{34\pm\sqrt{129}}{13}$ , so the first time is $1.74$ hours to $3$ significant figures.	6
Notes
At time $t$ , the position vectors are $(2t,t)$ for $A$ and $(10-t,-2+3t)$ for $B$ . Subtracting gives the displacement from $A$ to $B$ as $(10-3t,-2+2t)$ . A separation of $5$ km requires $(10-3t)^2+(-2+2t)^2=25$ , which simplifies to $13t^2-68t+79=0$ . Thus $t=[68\pm\sqrt{516}]/26=(34\pm\sqrt{129})/13$ . The first value is $1.741706\ldots$ , giving $1.74$ hours to $3$ significant figures.

(6 marks)

Q10

Tier 3 · Hard

10.

An aircraft must travel due north through a wind of

60

km h

^{-1}

due east. Its speed relative to the air is

250

km h

^{-1}

. Let

\mathbf{i}

point east and

\mathbf{j}

point north. Find the required air-velocity vector, the bearing on which the aircraft must head to the nearest

0.1^\circ

, and its ground speed to

3

significant figures.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	The required air-velocity vector is $-60\mathbf{i}+10\sqrt{589}\mathbf{j}$ km h $^{-1}$ . The required bearing is $346.1^\circ$ , and the ground speed is $243$ km h $^{-1}$ to $3$ significant figures.	6
Notes
To cancel the wind's eastward component, the aircraft's air velocity must have east component $-60$ . Write it as $-60\mathbf{i}+v\mathbf{j}$ with $v>0$ . Its magnitude is $250$ , so $60^2+v^2=250^2$ and $v=\sqrt{58900}=10\sqrt{589}$ . Adding the wind vector $60\mathbf{i}$ leaves the ground velocity $10\sqrt{589}\mathbf{j}$ , which is due north and has magnitude $242.693\ldots$ km h $^{-1}$ . The heading is $\tan^{-1}(60/(10\sqrt{589}))=13.886\ldots^\circ$ west of north, so the bearing is $360^\circ-13.886\ldots^\circ=346.1^\circ$ .

(6 marks)

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